Problem: $y=-3\tan^3(\sin(x))$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-9\sec^2(x)\sin^2(\tan(x))\cos(\tan(x))$ (Choice B) B $-9\sin^2(x)\cos(x)\sec^2(\sin^3(x))$ (Choice C) C $-9\tan^2(\sin(x))\sec^2(\sin(x))\cos(x)$ (Choice D) D $-9\tan^2(x)\sec^2(\sin(x))\cos(x)$
Explanation: $-3\tan^3(\sin(x))$ is a composition of three functions! Let... $u(x)=-3x^3$ $v(x)=\tan(x)$ $w(x)=\sin(x)$... then $y=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $\dfrac{dy}{dx}$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=-9x^2$ $v'(x)=\sec^2(x)$ $w'(x)=\cos(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={-9\tan(\sin(x))^2}\cdot{ \sec^2(\sin(x))}\cdot{\cos(x)} \\\\ &=-9\tan^2(\sin(x))\sec^2(\sin(x))\cos(x) \end{aligned}$ In conclusion: $\dfrac{dy}{dx}=-9\tan^2(\sin(x))\sec^2(\sin(x))\cos(x)$